HDU 1875 smooth Engineering

Solution:

It is still the smallest spanning tree. If the side length is less than 10 or greater than 1000, it will be skipped. Finally, check whether the Spanning Tree has n-1 edge.

#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <algorithm>#include <cmath>#include <vector>#include <queue>#include <map>#include <set>#include <stack>#define LL long long#define FOR(i,x,y) for(int i=x;i<=y;i++)#define rFOR(i,x,y) for(int i=x;i>=y;i--)using namespace std;const int maxn = 100 + 10;struct Point{double x , y;}P[maxn];int N , M;double get_dis(const Point& A , const Point& B){double x = A.x - B.x;double y = A.y - B.y;return sqrt(x * x + y * y);}struct Edge{int u;int v;double w;bool operator < (const Edge& rhs) const{return w < rhs.w;}}e[maxn * maxn];int f[maxn];int find(int x){return f[x] == x ? x : f[x] = find(f[x]);}void Kruskal(){sort(e , e + M);FOR(i,1,N) f[i] = i;double ans = 0 ; int n = 0;for(int i = 0 ; i < M ; i ++){if(e[i].w < 10 || e[i].w > 1000) continue;int x = find(e[i].u);int y = find(e[i].v);if(x != y){f[x] = y;ans += e[i].w;n++;}}if(n == N-1) printf("%.1lf\n",ans*100);else printf("oh!\n");}int main(){int T;scanf("%d",&T);while(T--){scanf("%d",&N);M = 0;FOR(i,1,N) scanf("%lf%lf",&P[i].x , &P[i].y);FOR(i,1,N){FOR(j,i+1,N){double dis = get_dis(P[i] , P[j]);e[M].u = i;e[M].v = j;e[M++].w = dis;}}//for(int i=0;i<M;i++) cout<<e[i].u<<' '<<e[i].v<<' '<<e[i].w<<endl;Kruskal();}return 0;}

HDU 1875 smooth Engineering